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Field problem

  1. Jul 1, 2007 #1
    help me its so hard

    working in the finite field Z_p show that the all the factors of polynomial x^{p^n}-x have degree "d" such that d|n.

  2. jcsd
  3. Jul 1, 2007 #2


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    use the binomial theorem and the fact that mod p, (x-a)^p = x^p-a^p.
  4. Jul 1, 2007 #3


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    Let a be a root of that polynomial. What does that tell you about Zp[a]?
    (or, did I mean to ask that the other way around?)

    Oh, even better: do you know the splitting field of that polynomial?
  5. Jul 1, 2007 #4
    I think you mean all irreducible factors.

    This can be avoided without splitting fields. But it is a little more lengthy.

    We can find [tex]\bar{\bold{Z}_p}[/tex] this is the algebraic closure of [tex]\bold{Z}[/tex]. Given [tex]p(x)[/tex] an irreducible factor we can choose [tex]\alpha \in \bar{\bold{Z}_p}[/tex] that is a zero. Now if we adjoin [tex]\alpha[/tex] to [tex]p(x)[/tex] we get a field [tex]\bold{Z}_p(\alpha)[/tex] with [tex][\bold{Z}_p(\alpha):\bold{Z}_p]=d[/tex] because that is the degree of [tex]p(x)[/tex]. Let [tex]F[/tex] be the set of all zeros in [tex]\bar{\bold{Z}_p}[/tex] to the polynomial [tex]x^{p^n}-x[/tex], this makes [tex]F[/tex] a field with [tex]p^n[/tex] elements. Since [tex]\bold{Z}_p \subset F[/tex] we have that [tex][F:\bold{Z}_p]=n[/tex]. Finally notice that [tex] \alpha \in F[/tex] since [tex]p(\alpha)=0[/tex] and [tex]p(x)|\left( x^{p^n}-x \right)[/tex]. This immediately implies that [tex]\bold{Z}_p (\alpha) \subseteq F[/tex]. Now we have succesfully established the fact that [tex]\bold{Z}_p\subseteq \bold{Z}_p(\alpha) \subseteq F[/tex]. Now using theorem for degrees of finite field extensions we have [tex][F:\bold{Z}_p(\alpha)][\bold{Z}_p(\alpha):\bold{Z}_p]=[F:\bold{Z}_p][/tex]. So [tex][F:\bold{Z}_p(\alpha)]\cdot d = n[/tex] which shows that [tex]d|n[/tex].
    Last edited: Jul 1, 2007
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